Jak można uprościć wyrażenie...

[morgoth404]

\(\displaystyle{ c= \cos 3^o \cdot \sqrt{1 + tg ^{2}3^o} +1}\)





Ps. pisząc samą 3, myślałem o 3 stopniach.

[soku11]

\(\displaystyle{ 1+\tan ^2 \alpha=
1+\frac{\sin^2 \alpha}{\cos^2\alpha}=
\frac{\sin^2\alpha+\cos^2\alpha}{\cos^2\alpha}=
\frac{1}{\cos^2\alpha}\\
\cos 3\cdot \sqrt{1+\tan^2 3}+1=
\cos 3\cdot \sqrt{\frac{1}{\cos^2 3}}+1=
\cos 3\cdot \frac{1}{|\cos 3|}+1=\ldots}\)

Pozdrawiam.

[Mersenne]

Chyba chodziło Ci o to:

\(\displaystyle{ \cos 3^{\circ} \cdot \sqrt{1+\tg^{2} 3^{\circ}}+1=\cos 3^{\circ} \cdot \sqrt{1+\frac{sin^{2} 3^{\circ}}{\cos^{2} 3^{\circ}}}+1=}\)
\(\displaystyle{ =\cos 3^{\circ} \cdot \sqrt{\frac{sin^{2} 3^{\circ}+\cos^{2} 3^{\circ}}{\cos^{2} 3^{\circ}}}+1=\cos 3^{\circ} \cdot \sqrt{\frac{1}{\cos^{2} 3^{\circ}}}+1=\cos 3^{\circ} \cdot \frac{1}{|cos 3^{\circ}|}+1=}\)
\(\displaystyle{ =\frac{\cos 3^{\circ}}{\cos 3^{\circ}}+1=1+1=2}\)